Circular and Satellite Motion Legacy Problem #11 Guided Solution
Problem*
Alexis is in her Toyota Camry and trying to make a turn off an expressway at 19.0 m/s. The turning radius of the level curve is 35.0 m. Her car has a mass of 1240 kg. Determine the acceleration, net force and minimum value of the coefficient of friction which is required to keep the car on the road.
Audio Guided Solution
Alexis is in her Toyota and trying to make a turn off of an expressway. We're given her speed, V equal 19.0 meters per second, and we're given the radius of the curve along which she's traveling, R equal 35.0 meters. Her mass, the mass of her car is 1240 kilograms. We're asked to calculate three things, A, F net, and the value of mu, the coefficient of friction. Mu is in the equation, F friction equal mu F norm. My solution to the problem centers around a free body diagram. It's a level turn, and so what I know about the gravity force is it's straight down and the normal force is straight up. So I draw a box representing Alexis' car. I draw it narrowed down, I call it F grav, I say it's equal mg, and I draw the up force, normal force. Since they will balance each other, I know that F norm equal mg also. So the last force is the friction force. It's the force towards the center of the turn. It's the net force, F friction is, and so I can say that F net equal F friction. Now my net force is always ma, and my friction force is mu times F norm. So I can rewrite my F net equal F friction equation as ma equal mu F norm. Now thinking a little further about this mu times F norm, I could think F norm is just mg, and I could rewrite my equation now as ma equal mu times mg. Now you'll notice the m's cancel, and I have a mu in the equation. So I can rearrange this equation to say mu equal a over g. If I can only find the acceleration, I can put it into this equation, and I can solve for the value of mu, the coefficient of friction. That's my strategy. So how do I calculate the a? Well I notice that the speed is given, and I notice the radius of the turn is given. So a equals speed squared over r. a equals 19 squared over 35. When I do the math, I get 10.3143. Now I can take the 10.3143, and I can say mu is equal to this value a divided by 9.8. When I do that, I get about 1.0525, which is a relatively large coefficient of friction. If I were a betting person, I'd bet that Alexis isn't going to make it around this level turn. And this explains one reason why we have speed limits upon turns, and another reason why we usually bank the turns.
Solution
1.05
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities in an organized manner. Equate given values to the symbols used to represent the corresponding quantity - e.g., \(\descriptive{m}{m,mass} = 61.7\unit{kg}\), \(\descriptive{v}{v,velocity} = 18.5 \unit{\meter\per\second}\), \(\descriptive{R}{R,radius} = 30.9\unit{m}\), \(F_\text{norm} = \colorbox{gray}{Unknown}\).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
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